Answer:
x < -12 or x > 16.
Explanation:
We have:
Ix - 2|+ 3 > 17
To solve this equation, we first need to isolate the absolute value term. We do this by subtracting 3 from both sides of the equation, giving us:
Ix - 2|+ 3 -3> 17-3
Ix - 2| > 14
Now, we need to consider two cases: the case where x - 2 is positive, and the case where x - 2 is negative.
- Case 1: x - 2 is positive.
In this case, we can remove the absolute value bars, because the expression inside the bars is already positive. So, we have:
x - 2 > 14
Adding both side by 2.
x-2+2 > 14+2
x > 16
This means that x > 16.
- Case 2: x - 2 is negative
In this case, we can write the expression inside the absolute value bars as its negative equivalent. So, we have:
|x - 2| = -(x - 2)
Now, we can remove the absolute value bars, because the expression inside the bars is now negative. So, we have:
-(x - 2) > 14
opening bracket
-x +2 > 14
Adding x on both sides
-x+x+2 > 14+x
2 > 14+ x
Subtracting 14 on both sides
2-14> x
x <-12
This means that x < -12.
Combining the solutions from both cases, we get that x < -12 or x > 16.
Also written as : - 12 > x > 16
